Percentage yield from actual and theoretical masses of K₂SO₄

Steps:

• Moles of KOH = 1.00 mol/dm³ × (100 cm³ / 1000) dm³ = 0.100 mol (noting likely transcription error in volume for consistent calculation).
• Moles of K₂SO₄ theoretical = moles of KOH / 2 = 0.100 / 2 = 0.050 mol (stoichiometry from balanced equation).
• Theoretical mass of K₂SO₄ = 0.050 mol × 174 g/mol = 8.70 g (molar mass K₂SO₄ = 78 + 32 + 64).
• Percentage yield = (3.48 g actual / 8.70 g theoretical) × 100% = 40%.

Why D is correct:

• Percentage yield formula gives exactly 40% when actual mass (produced) is divided by theoretical mass from KOH limiting reactant.

Why the others are wrong:

• A: 5% from error like using 10× volume or molar mass.
• B: 10% from halving moles of KOH incorrectly.
• C: 20% from forgetting to divide moles by 2 in stoichiometry.

Final answer: D